T 1. The value function v(x) is the expected net gain when using the optimal stopping time starting at state x:

Size: px

Start display at page:

Download "T 1. The value function v(x) is the expected net gain when using the optimal stopping time starting at state x:"

Shannon Hill
5 years ago
Views:

1 108 OPTIMAL STOPPING TIME 4.4. Cost functions. The cost function g(x) gives the price you must pay to continue from state x. If T is your stopping time then X T is your stopping state and f(x T ) is your payoff. But your cost to play to that point was T 1 cost = g(x 0 ) + g(x 1 ) + + g(x T 1 ) = g(x j ) So, your net gain is T 1 net = f(x T ) g(x j ) The value function v(x) is the expected net gain when using the optimal stopping time starting at state x: It satisfies the equation: j=0 j=0 T 1 v(x) = E(f(X T ) g(x j ) X 0 = x) j=0 v(x) = max(f(x), (P v)(x) ) Proof: In state x you should either stop or continue. (1) If you stop you get f(x). () If you continuous and use the optimal strategy after that then you get v(y) with probability p(x, y) but you have to pay g(x). So, you would expect to get p(x, y)v(y) y x You should pick the one which gives you a higher expected net. So, v(x) is the maximum of these two numbers nonstochastic case. The first example was a dice problem. You toss two dice and let x = the sum. Then x 1 with probabilities indicated below. p = /36 x = f(x) = g(x) = There was a question of what should be g(7). It does not make sense to talk about the cost of continuing from 7 if you are not allowed to continue. So, I decided that, in order to have a well defined function (g(x) needs to be defined for every x S in order for g to be a function),

2 MATH 56A SPRING 008 STOCHASTIC PROCESSES 109 we should allow the player to pay max f(x) = 1 to continue from 7. It doesn t make sense to pay 1 to play a game where the maximum gain is 1. So, this has the effect of making 7 recurrent. The problem is to find the value function v(x) and the optimal strategy (the formula for T ). I pointed out that this Markov chain is actually not stochastic in the sense that the probabilities do not change with time. This implies that the value function v(x) which is a vector with 10 unknown coordinates (11 coordinates of which we know only v(7) = 0): v = (v(), v(3),, v(6), v(7) = 0, v(8),, v(1) is determined by one number E = the expected payoff if you continue Then, your expected net if you continue is E so And E is given in terms of v by: v(x) = max(f(x), E ) if x 7. E = x 7 p x v(x) When you do the iteration algorithm you compute and you get a sequence of numbers All you need is the single number E n = x 7 p x u n (x) E 1, E, E 3, E = E = lim n E n.

3 110 OPTIMAL STOPPING TIME No cost First I did this in the no cost case. When n = 1 you take the most optimistic view: Hope to get x = 1. But you have a probability p 7 = 6/36 = 1/6 of getting x = 7 and losing. So, E 1 = x 7 p x u 1 (x) = x 7 p x 1 = (5/6)1 = 10. Then u (x) = max(f(x), E) if x 7 (But make sure to put u (7) = 0): x = f(x) = E = u (x) = If you take the average value of u (x) you get E : Repeating this process you get: E = x 7 p x u (x) = E 3 = E 4 = E 5 = E 6 = E 5 = Once you realize that E is somewhere between 6 and 7 you know the winning strategy: You need to continue if you get 6 or less and stop if you get 8 or more. So, which makes v(x) = (E, E, E, E, E, 0, 8, 9, 10, 11, 1) E = 1 (E + E + 3E + 4E + 5E + 5(8) + 4(9) + 3(10) + (11) + 1) 36 = 1 (15E + 140) 36 36E = 15E E = 140/1 = 0/3 = 6 3 So, the value function is the vector: v = ( 0, 0, 0, 0, 0, 0, 8, 9, 10, 11, 1)

4 MATH 56A SPRING 008 STOCHASTIC PROCESSES 111 With cost The iteration algorithm starts with (I don t remember what I said but you have to remember that u n (x) f(x) all the time): 0 if x = 7 f(x) if f(x) max f(y) This gives: u 1 (x) = max f(y) }{{} hope for best }{{} cost otherwise u 1 = (10, 10, 10, 10, 11, 0, 11, 11, 11, 11, 1) The average of these numbers is: E 1 = x 7 p x u 1 (x) = Then u (x) = max(f(x), E 1 ) u = (6.917, 6.917, 6.917, 6.917, 7.917, 0, 8, 9, 10, 11, 1) with average E = p x u (x) = x 7 E 3 = E 4 = E 10 = E 11 = We just needed to know that E is between 5 and 6. This tells us that the optimal strategy is to continue if you get or 3 and stop if you get 4 or more. After you determine the optimal strategy, you can find the exact value of both E and the value function v(x). First you find v(x) in terms of E: v(x) = (E, E, 4, 5, 6, 0, 8, 9, 10, 11, 1) The average of these numbers is E. So, E = (E )(3/36) + 0/36 E = 196/33 = 5 31 = 5 93 = v(x) = (3 31, 3 31, 4, 5, 6, 0, 8, 9, 10, 11, 1) 33 33

5 11 OPTIMAL STOPPING TIME random walk. with absorbing walls. In the general (stochastic) case the value function is the solution of the equation: v(x) = max(f(x), p(x, y)v(y) g(x)) y }{{} v(x 1)+v(x+1) In the case of the random walk, this part is what we have before. So, v(x) is the smallest function so that v(x) v(x) f(x) and v(x 1) + v(x + 1) Example 4.8. Suppose the payoff and cost functions are: states x = f(x) = g(x) = = w(3) 11 = f(3) 15 v(0)+v() 5 1 f(1) = 5 w(1) = 4 1 In the graph, the thin lines gives the convex hull of the function f(x). This would be the answer if there were no cost. Since the cost is 1, we have to go one step below the average. I called this function w(x): w(x) := f(x 1) + f(x + 1) Since v(x) f(x), it must also be w(x): v(x) v(x 1) + v(x + 1) g(x) f(x 1) + f(x + 1) g(x) = w(x)

6 MATH 56A SPRING 008 STOCHASTIC PROCESSES 113 This is a simple case in which the gaps have length. So, we can just compare f(x) and w(x) to get the value function v(x). If the gap is more than two, the equation becomes more complicated. I think I forgot to say this: For the iteration algorithm we can start with the value function that we know how to calculate when there is no cost: u 1 (x) = (0, 5 1, 11, 13, 15, 0) Then ( u (x) = max f(x), u ) 1(x 1) + u 1 (x + 1) u = (0, 5, 11, 1, 15, 0) If you do it again, you get the same thing: u 3 = u. So, this is also equal to the value function: v = (0, 5, 11, 1, 15, 0) So, the optimal strategy is to continue when x = 3 (since that is the only point where v(x) > f(x)) but stop at any other point.

MATH 56A SPRING 2008 STOCHASTIC PROCESSES

MATH 56A SPRING 2008 STOCHASTIC PROCESSES MATH 56A SPRING 008 STOCHASTIC PROCESSES KIYOSHI IGUSA Contents 4. Optimal Stopping Time 95 4.1. Definitions 95 4.. The basic problem 95 4.3. Solutions to basic problem 97 4.4. Cost functions 101 4.5.